f(g(x))
= f(4/x + 4)
= 1/(4/x+4 - 4)
= 1/(4/x)
= x/4
there is no restriction on this final result, BUT
in the original g(x) = 4/x + 4, there was the restriction of x ≠ 0
in my view that restriction should carry over to the final function.
eg
g(3) = 4/3 + 4 = 16/3
f(16/3) = 1/(16/3 - 4) = 3/4
and according to my result f(g(3)) = 3/4, the correct result
g(0) is undefined
f(undefined) = 1/(undefined - 4), but my final function would give us 0
Use the given functions to find f(g(x)), and give the restrictions on x.
f(x) = 1 / (x-4)
g(x) = (4/x) + 4
What are the restrictions of x? How do you look for that?
5 answers
extra credit
f(x) = 1/(x-4) is undefined at x = 4
g(x) = 4/x + 4 is undefined at x = 0
so, why is f(g(x)) not undefined at x=4?
f(x) = 1/(x-4) is undefined at x = 4
g(x) = 4/x + 4 is undefined at x = 0
so, why is f(g(x)) not undefined at x=4?
Yup, oobleck is right
so, would be have x ≠ 0,4 ??
so, would be have x ≠ 0,4 ??
So both 0 and 4 would be restrictions?
4 coming from the f(x) and
0 coming from the g(x)
4 coming from the f(x) and
0 coming from the g(x)
note that g(x) is never equal to 4